Timeline for Case study: what does it take to formulate and prove Quillen's small object argument in ZFC?
Current License: CC BY-SA 4.0
17 events
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| Feb 6, 2021 at 3:51 | comment | added | Tim Campion | I think my imagination was a bit weak earlier. If the regularity assumption is weakened, then sure, we'll get a weak "external" version of the theorem where the presentability ranks are weakened. But so long as $V_\kappa$ models enough of ZFC, there will also be an "internal" version of the theorem where we use as many "internal to $V_\kappa$" concpets as possible... I'm unclear on how it will all play out. | |
| Jan 29, 2021 at 3:21 | comment | added | Tim Campion | Is $\Sigma_1$-replacement equivalent over $ZC$ to "every set is bijective to an ordinal"? | |
| Jan 29, 2021 at 3:07 | comment | added | Noah Schweber | Yes, that's right - and it's exactly $\Sigma_1$ replacement that's needed. (As an unrelated aside, this is a key point in generalized computability theory, where that's the amount of replacement we have at admissible ordinals and is crucial for developing basic computability-theoretic results - although that doesn't end the story.) | |
| Jan 29, 2021 at 3:05 | comment | added | Tim Campion | And although this might seem like a different issue than the one I originally asked about, it really can still be seen as a question about replacement, since if $V_\kappa$ models some pretty weak replacement ($\Sigma_1$ or something), then it models that every well-order is isomorphic to an ordinal, and hence $\kappa$ must be a $\beth$-fixed-point anyway. | |
| Jan 29, 2021 at 3:01 | comment | added | Tim Campion | In the discussion at Peter Scholze's question, it emerged that if $V_\kappa$ models $\Sigma_n$-replacement, then typically $\kappa$ will be very far from regular, leading me to think that the "canonical" approach would be to weaken the regularity assumption. But if we do that, then I can't imagine that the theorem can possibly be proven for sets of morphisms $I$ whose domains and codomains have presentability rank $\geq cf(\kappa)$. So the resulting theorem will be weaker. Perhaps it would still be applicable "in practice". Ugh. This stuff is annoying. | |
| Jan 29, 2021 at 2:54 | comment | added | Noah Schweber | Yes, that's right. | |
| Jan 29, 2021 at 2:53 | comment | added | Tim Campion | Sorry, I meant that we don't get an "if and only if". If $\kappa$ isn't a $\beth$-fixed-point, then there are sets in $V_\kappa$ of cardinality $\geq \kappa$, right? | |
| Jan 29, 2021 at 2:52 | comment | added | Noah Schweber | "If $\kappa$ isn't a $\beth$-fixed-point, then we can't just say that every set of cardinality $<\kappa$ is bijective with a set in $V_\kappa$" Every set of cardinality $<\kappa$ is in bijection with some ordinal $<\kappa$ which is an element of $V_\kappa$, so $\beth$-fixed-pointness isn't needed here at all. Did you mean something else? | |
| Jan 29, 2021 at 2:49 | comment | added | Tim Campion | I've now spent way too much time thinking about the necessary preliminaries and I'm stuck at the following conundrum. ZFC won't give us $\kappa$ which is both regular and a $\beth$-fixed-point. If $\kappa$ isn't regular, then we can't run transfinite induction for arbitrary lengths $\sigma <\kappa$ and we'll get a weaker theorem. If $\kappa$ isn't a $\beth$-fixed-point, then we can't just say that every set of cardinality $<\kappa$ is bijective with a set in $V_\kappa$. We'll have to cook up more elaborate ways of controlling "size". I'm starting to think this would be a nontrivial endeavor. | |
| Jan 29, 2021 at 2:25 | history | rollback | Tim Campion |
Rollback to Revision 1
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| Jan 29, 2021 at 2:24 | history | edited | Tim Campion | CC BY-SA 4.0 |
I got this far and realized that the $\beth$-fixed-point assumption is going to be too strong if we're going to eventually assume regularity. I am saving this and will immediately rollback just for my own records.
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| Jan 28, 2021 at 21:22 | comment | added | Tim Campion | @NoahSchweber I'm not sure what "very closed" means, but if $\theta \ll \theta$ doesn't imply that $\theta$ is strongly inaccessible, then most likely my use of it in the estimate is actually in error. I should probably work out the estimate in more detail. | |
| Jan 28, 2021 at 21:19 | comment | added | Noah Schweber | So I don't immediately see why $\kappa$ has to be a limit cardinal. | |
| Jan 28, 2021 at 21:17 | comment | added | Noah Schweber | "Regular limit cardinal" is just weakly inaccessible and already $\mathsf{ZFC}$ doesn't prove the existence of those. However, I don't think what you've written corresponds to that. In particular, consider what happens if $\kappa=\theta^+$ for some "very closed" $\theta$ (since successors are regular): we might as well take $\mu=\theta$, but then $\rho=\theta$ works since we only consider $\mu'<\mu$ and $\rho'<\rho$ and $\theta$ is "very closed." (Specifically, given your favorite $\lambda$ take $\theta$ to be the limit of the sequence $\theta_0=\lambda$, $\theta_{i+1}=\theta_i^{\theta_i}$.) | |
| Jan 28, 2021 at 21:10 | comment | added | Tim Campion | Hang on -- after running through my limited knowledge of cardinal arithmetic I'm thinking that perhaps the estimates I gave are not good enough. I believe I've been told it's consistent that every regular limit cardinal is strongly inaccessible (I think this always holds in L?). The conditions I gave imply that $\kappa$ is a regular limit cardinal. Ergo, ZFC does not prove that there exists a $\kappa$ with the properties I described. Probably I've just missed how to get a better estimate. | |
| S Jan 28, 2021 at 21:02 | history | answered | Tim Campion | CC BY-SA 4.0 | |
| S Jan 28, 2021 at 21:02 | history | made wiki | Post Made Community Wiki by Tim Campion |